If the area bounded by the curve 2y² = 3x, lines x + y = 3, y = 0 and outside the circle (x - 3)² +

Q: Why did we integrate with respect to $y$ instead of $x$? Isn't $y=f(x)$ the standard way?

Great question! While $y=f(x)$ is standard, it often leads to complex piecewise functions when boundaries change. By integrating with respect to $y$, we treat the parabola and line as functions of $y$, which simplifies the integrand to a single polynomial expression. Always choose the path of least resistance!

Q: How do I know the circle sector is exactly $\frac{\pi}{4}$?

Look at the line $x+y=3$. Its slope is $-1$, which corresponds to an angle of $135^\circ$ with the positive x-axis, or $45^\circ$ ($\frac{\pi}{4}$ radians) with the negative x-axis. Since the circle is centered at $(3,0)$, the sector trapped between the x-axis and this line is exactly $\frac{\pi}{4}$.

Q: Why subtract the sector area? Does the circle overlap the entire region?

The problem asks for the area 'outside the circle'. By calculating the total area bounded by the curves first, we find the 'gross' area. We then identify the 'net' area by removing the portion of the circle that lies within those boundaries. The sector is the exact geometric shape of that overlap.

Q: Is it possible to solve this using polar coordinates?

Technically yes, but it would be like using a sledgehammer to crack a nut! The boundaries are linear and parabolic; Cartesian coordinates are much more natural here. Stick to the simplest tool that gets the job done.

Q: I got a negative value for the area. What did I do wrong?

Area must always be positive. If you get a negative result, you likely swapped the upper and lower boundaries in your integral. Remember: $Area = \int (x_{right} - x_{left}) dy$. Ensure your 'right' function is indeed to the right of the 'left' function!

Question

If the area bounded by the curve $2y^2 = 3x$, lines $x + y = 3, y = 0$ and outside the circle $(x - 3)^2 + y^2 = 2$ is $A$, then $4(\pi + 4A)$ is equal to _________.

Sigma Percentile · Accepted Answer

### Analyzing the Terrain

We begin by defining our boundaries. We have the parabola $2y^2 = 3x$, which is equivalent to $x = \frac{2y^2}{3}$. This represents a standard parabola opening to the right.

We also have the line $x + y = 3$, which can be written as $x = 3 - y$, and the $x$-axis, defined by $y = 0$.

To find the intersection of the parabola and the line, we set the $x$-values equal:
$$\frac{2y^2}{3} = 3 - y$$

Rearranging this yields the quadratic equation $2y^2 + 3y - 9 = 0$. Factoring this expression, we obtain:
$$(2y - 3)(y + 3) = 0$$

Since our region is bounded by $y=0$, we focus on the positive root, $y = \frac{3}{2}$. This value serves as our upper limit for integration.

### The Power of Integration

To calculate the total area $A_{total}$ trapped between the line and the parabola, we integrate with respect to $y$. This approach avoids splitting the area into multiple parts.

The integral is defined as the difference between the 'right' function and the 'left' function:
$$A_{total} = \int_0^{\frac{3}{2}} \left( (3 - y) - \frac{2y^2}{3} ight) dy$$

Performing the integration term by term, we get:
$$\left[ 3y - \frac{y^2}{2} - \frac{2y^3}{9} ight]_0^{\frac{3}{2}}$$

Substituting the upper limit $y = \frac{3}{2}$, we calculate:
$$3\left(\frac{3}{2}ight) - \frac{1}{2}\left(\frac{9}{4}ight) - \frac{2}{9}\left(\frac{27}{8}ight) = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{21}{8}$$

### The Circle's Intrusion

The problem requires us to exclude the area inside the circle $(x - 3)^2 + y^2 = 2$. This circle is centered at $(3, 0)$ with a radius $r = \sqrt{2}$.

The line $x + y = 3$ passes through the center of the circle $(3, 0)$. The angle between the $x$-axis and this line is $\frac{\pi}{4}$.

Consequently, the circle carves out a sector of area defined by $\frac{1}{2}r^2	heta$. Substituting our values:
$$Area_{sector} = \frac{1}{2}(\sqrt{2})^2\left(\frac{\pi}{4}ight) = \frac{\pi}{4}$$

### Final Synthesis

Our net area $A$ is the total area minus the sector area:
$$A = A_{total} - Area_{sector} = \frac{21}{8} - \frac{\pi}{4}$$

The problem asks us to evaluate the expression $4(\pi + 4A)$. First, we calculate $4A$:
$$4A = 4\left(\frac{21}{8} - \frac{\pi}{4}ight) = \frac{21}{2} - \pi$$

Adding $\pi$ to this result, we obtain:
$$\pi + \left(\frac{21}{2} - \piight) = \frac{21}{2}$$

Finally, multiplying by $4$ gives the result:
$$4 	imes \frac{21}{2} = 42$$

The final answer is **42**.

Animated Solution for Mathematics - Definite Integration: If the area bounded by the curve $2 y^{2} = 3 x$ , lines $x + y = 3, y = 0$ and outside the circle $(x - 3)^{2} + y^{2} = 2$ is $A$ , then $4 (π + 4 A)$ is equal to _________.

Enter Numerical Value:

Visualized Solution

Identify the Curves

Intersection of Parabola and Line

Integral for Total Area

Evaluate $A_{t o t a l}$

The Circle Constraint

Area of the Sector

Calculate Area $A$

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Terrain

The Power of Integration

The Circle's Intrusion

Final Synthesis

Similar Questions

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If the area of the region ${(x, y) : ∣ x^{2} - 2∣ \leq y \leq x}$ is $A$ , then $6 A + 162$ is equal to ______________

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Area of the region ${(x, y) : x^{2} + (y - 2)^{2} \leq 4, x^{2} \geq 2 y}$ is

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is:

Enter Numerical Value:

Visualized Solution

Identify the Curves

Intersection of Parabola and Line

Integral for Total Area

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Atotal​

The Circle Constraint

Area of the Sector

Calculate Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Terrain

The Power of Integration

The Circle's Intrusion

Final Synthesis

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